Solving $x^2 - 3x > 0$: Interval Notation Explained

by Alex Johnson 52 views

Understanding how to solve inequalities is a fundamental skill in mathematics, particularly in algebra and calculus. In this article, we will delve into the process of solving the quadratic inequality x2βˆ’3x>0x^2 - 3x > 0 and expressing the solution using interval notation. This step-by-step guide will help you grasp the concepts and techniques involved, ensuring you can confidently tackle similar problems. Let’s break it down and make it easy to understand.

Understanding Quadratic Inequalities

When dealing with quadratic inequalities like x2βˆ’3x>0x^2 - 3x > 0, the primary goal is to find the range(s) of xx values that satisfy the inequality. Quadratic inequalities differ from linear inequalities due to the presence of the x2x^2 term, which introduces a curve (parabola) when graphed. This curvature means that the solution often consists of intervals rather than single values or simple ranges. To effectively solve these inequalities, we need to find the critical points where the quadratic expression equals zero, as these points will delineate the intervals where the expression is either positive or negative.

To solve the quadratic inequality, it's crucial to understand the behavior of quadratic functions. A quadratic function generally takes the form ax2+bx+cax^2 + bx + c, and its graph is a parabola. The parabola opens upwards if a>0a > 0 and downwards if a<0a < 0. The roots of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 are the points where the parabola intersects the x-axis. These roots, also known as critical points, divide the number line into intervals. In each interval, the quadratic expression will either be consistently positive or consistently negative. By determining the sign of the expression in each interval, we can identify the regions where the inequality holds true. This involves testing values from each interval in the original inequality to see if they satisfy the condition.

Step-by-Step Solution for x2βˆ’3x>0x^2 - 3x > 0

Let’s go through the process step-by-step to make sure we fully understand the solution. By understanding each step, you can better apply these concepts to other inequalities.

Step 1: Factor the Quadratic Expression

The first crucial step in solving the inequality x2βˆ’3x>0x^2 - 3x > 0 is to factor the quadratic expression. Factoring simplifies the expression and allows us to identify the critical points more easily. By factoring, we are essentially rewriting the quadratic expression as a product of simpler expressions, which makes it easier to determine when the entire expression is greater than zero.

To factor x2βˆ’3xx^2 - 3x, we look for common factors in both terms. In this case, both terms have an xx in common. Factoring out xx gives us:

x2βˆ’3x=x(xβˆ’3)x^2 - 3x = x(x - 3)

Now, our inequality looks like this:

x(xβˆ’3)>0x(x - 3) > 0

This factored form is much easier to work with as it allows us to see the values of xx that make the expression equal to zero, which are the critical points for solving the inequality.

Step 2: Find the Critical Points

The critical points are the values of xx that make the expression equal to zero. These points are crucial because they divide the number line into intervals where the expression is either positive or negative. Essentially, we are finding the points where the parabola, represented by the quadratic expression, intersects the x-axis.

To find these points, we set each factor equal to zero and solve for xx:

  1. x=0x = 0
  2. xβˆ’3=0=>x=3x - 3 = 0 => x = 3

So, the critical points are x=0x = 0 and x=3x = 3. These points are where the expression x(xβˆ’3)x(x - 3) changes its sign. Now, we know the boundaries of our intervals, and we can move on to testing these intervals to determine where the inequality x(xβˆ’3)>0x(x - 3) > 0 holds true.

Step 3: Create Intervals Using Critical Points

The critical points we found, x=0x = 0 and x=3x = 3, divide the number line into three intervals. These intervals are the ranges of xx values that we need to test to see if they satisfy the inequality. By breaking the number line into these intervals, we can analyze the behavior of the quadratic expression in each region.

The three intervals are:

  1. (βˆ’βˆž,0)(-\infty, 0)
  2. (0,3)(0, 3)
  3. (3,∞)(3, \infty)

Each interval represents a range of xx values. For instance, (βˆ’βˆž,0)(-\infty, 0) includes all numbers less than 0, (0,3)(0, 3) includes numbers between 0 and 3, and (3,∞)(3, \infty) includes all numbers greater than 3. These intervals are essential for determining the solution to our inequality because the sign of the quadratic expression x(xβˆ’3)x(x - 3) will remain constant within each interval. This allows us to test one value from each interval to determine the sign of the entire interval.

Step 4: Test Values in Each Interval

To determine whether the inequality x(xβˆ’3)>0x(x - 3) > 0 is satisfied in each interval, we need to test a value from each interval in the expression. This process involves selecting a test point within each interval and plugging it into the factored inequality x(xβˆ’3)x(x - 3). By evaluating the expression at these test points, we can determine the sign of the expression within each interval.

  1. Interval (βˆ’βˆž,0)(-\infty, 0)

    • Choose a test value: Let’s pick x=βˆ’1x = -1
    • Substitute into the inequality: (βˆ’1)((βˆ’1)βˆ’3)=(βˆ’1)(βˆ’4)=4(-1)((-1) - 3) = (-1)(-4) = 4
    • Since 4>04 > 0, the inequality is satisfied in this interval.
  2. Interval (0,3)(0, 3)

    • Choose a test value: Let’s pick x=1x = 1
    • Substitute into the inequality: (1)((1)βˆ’3)=(1)(βˆ’2)=βˆ’2(1)((1) - 3) = (1)(-2) = -2
    • Since βˆ’2β‰―0-2 \ngtr 0, the inequality is not satisfied in this interval.
  3. Interval (3,∞)(3, \infty)

    • Choose a test value: Let’s pick x=4x = 4
    • Substitute into the inequality: (4)((4)βˆ’3)=(4)(1)=4(4)((4) - 3) = (4)(1) = 4
    • Since 4>04 > 0, the inequality is satisfied in this interval.

By testing these values, we have effectively mapped out the regions where the inequality x(xβˆ’3)>0x(x - 3) > 0 holds. The intervals where the expression is positive are the intervals that form our solution.

Step 5: Write the Solution in Interval Notation

Now that we have identified the intervals where the inequality x(xβˆ’3)>0x(x - 3) > 0 is satisfied, we can express the solution in interval notation. Interval notation is a concise way to represent sets of numbers, using parentheses and brackets to indicate whether endpoints are included or excluded.

From our testing in Step 4, we found that the inequality is satisfied in the intervals (βˆ’βˆž,0)(-\infty, 0) and (3,∞)(3, \infty). Since the inequality is strictly greater than zero (>>), we do not include the critical points 00 and 33 in the solution. This is why we use parentheses instead of brackets in the interval notation.

To combine these intervals into a single solution, we use the union symbol, βˆͺ\cup. The union symbol means that the solution includes all values in either interval.

Therefore, the solution to the inequality x2βˆ’3x>0x^2 - 3x > 0 in interval notation is:

(βˆ’βˆž,0)βˆͺ(3,∞)(-\infty, 0) \cup (3, \infty)

This notation indicates that the solution includes all real numbers less than 0 and all real numbers greater than 3. By using interval notation, we can clearly and accurately represent the solution set of the inequality.

Conclusion

Solving quadratic inequalities involves several key steps: factoring the quadratic expression, finding critical points, creating intervals, testing values within those intervals, and finally, expressing the solution in interval notation. By following these steps, we can effectively solve inequalities like x2βˆ’3x>0x^2 - 3x > 0 and understand the range of values that satisfy the given condition. This skill is essential for more advanced topics in mathematics, making it a crucial concept to master.

If you're interested in learning more about inequalities and other algebraic concepts, you might find helpful resources on websites like Khan Academy's Algebra Section.